Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class Main {
    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer buffer;

    private static class Point {
        int x,y;
        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    public static void solve() {
        int n = ni(), m = ni();
        int[][]g = new int[n][m];
        for (int i=0;i<n;i++) {
            for (int j=0;j<m;j++) {
                g[i][j] = ni();
            }
        }

        ArrayList<String>res = new ArrayList<>();

        if (m>=n) {

            for (int row = 0; row < n; row++) {
                int min = g[row][0];
                for (int col = 1; col < m; col++) min = Math.min(min, g[row][col]);
                for (int x = 0; x < min; x++) res.add("row " + (row + 1));
                for (int c = 0; c < m; c++) g[row][c] -= min;
            }

            for (int col = 0; col < m; col++) {
                int min = g[0][col];
                for (int row = 1; row < n; row++) min = Math.min(min, g[row][col]);
                for (int x = 0; x < min; x++) res.add("col " + (col + 1));
                for (int r = 0; r < n; r++) g[r][col] -= min;
            }
        } else {
            for (int col = 0; col < m; col++) {
                int min = g[0][col];
                for (int row = 1; row < n; row++) min = Math.min(min, g[row][col]);
                for (int x = 0; x < min; x++) res.add("col " + (col + 1));
                for (int r = 0; r < n; r++) g[r][col] -= min;
            }

            for (int row = 0; row < n; row++) {
                int min = g[row][0];
                for (int col = 1; col < m; col++) min = Math.min(min, g[row][col]);
                for (int x = 0; x < min; x++) res.add("row " + (row + 1));
                for (int c = 0; c < m; c++) g[row][c] -= min;
            }
        }

        boolean empty = true;
        for (int i=0;i<n;i++) {
            for (int j=0;j<m;j++) {
                if(g[i][j]>0) {
                    empty = false;
                    break;
                }
            }
            if (!empty) break;
        }

        if (empty) {
            out.println(res.size());
            for (String val: res) out.println(val);

        } else out.println(-1);
    }


    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    static String next() {
        while (buffer == null || !buffer.hasMoreElements()) {
            try {
                buffer = new StringTokenizer(in.readLine());
            } catch (IOException e) {
                //
            }
        }
        return buffer.nextToken();
    }

    static int ni() {
        return Integer.parseInt(next());
    }

    static long nl() {
        return Long.parseLong(next());
    }

    static double nd() {
        return Double.parseDouble(next());
    }

    static String ns() {
        return next();
    }

    static int[] ni(int n) {
        int[] res = new int[n];
        for (int i = 0; i < n; i++)
            res[i] = Integer.parseInt(next());
        return res;
    }

    static long[] nl(int n) {
        long[] res = new long[n];
        for (int i = 0; i < n; i++)
            res[i] = Long.parseLong(next());
        return res;
    }

    static double[] nd(int n) {
        double[] res = new double[n];
        for (int i = 0; i < n; i++)
            res[i] = Double.parseDouble(next());
        return res;
    }

    static String[] ns(int n) {
        String[] res = new String[n];
        for (int i = 0; i < n; i++)
            res[i] = next();
        return res;
    }


}